Question: Relativisitic Fermi gas ([1], pr 9.3)

Consider a relativisitic gas of particles of spin obeying Fermi statistics, enclosed in volume , at absolute zero. The energy-momentum relation is

where , and is the rest mass.

Find the Fermi energy at density .

With the pressure defined as the average force per unit area exerted on a perfectly-reflecting wall of the container.

Set up expressions for this in the form of an integral.

Define the internal energy as the average .
Set up expressions for this in the form of an integral.

Show that at low densities, and at high densities. State the criteria for low and high densities.

There may exist a gas of neutrinos (and/or antineutrinos) in the cosmos. (Neutrinos are massless Fermions of spin .) Calculate the Fermi energy (in eV) of such a gas, assuming a density of one particle per .

Attempt exact evaluation of the various integrals.

Answer

We’ve found [3] that the density of states associated with a 3D relativisitic system is

For a given density , we can find the Fermi energy in the same way as we did for the non-relativisitic energies, with the exception that we have to integrate from a lowest energy of instead of (the energy at ). That is

Solving for we have

We’ll see the constant factor above a number of times below and designate it

so that the Fermi energy is

For the pressure calculation, let’s suppose that we have a configuration with a plane in the orientation as in fig. 1.1.

Fig 1.1: Pressure against x,y oriented plane

It’s argued in [4] section 6.4 that the pressure for such a configuration is

where is the number density and is a normalized distribution function for the velocities. The velocity and momentum components are related by the Hamiltonian equations. From the Hamiltonian eq. 1.1 we find \footnote{ Observe that by squaring and summing one can show that this is equivalent to the standard relativisitic momentum .} (for the x-component which is representative)

For we can summarize these velocity-momentum relationships as

Should we attempt to calculate the pressure with this parameterization of the velocity space we end up with convergence problems, and can’t express the results in terms of . Let’s try instead with a distribution over momentum space

Here the momenta have been scaled to have units of energy since we want to express this integral in terms of energy in the end. Our normalized distribution function is

but before evaluating anything, we first want to change our integration variable from momentum to energy. In spherical coordinates our volume element takes the form

Implicit derivatives of

gives us

Our momentum volume element becomes

For our distribution function, we can now write

where is determined by the requirement

The z component of our momentum can be written in spherical coordinates as

Noting that

all the bits come together as

Letting , this is

We could conceivable expand the numerators of each of these integrals in power series, which could then be evaluated as a sum of terms.

Note that above the Fermi energy also has an integral representation

or

Observe that we can use this result to remove the dependence of pressure on this constant

Now for the average energy difference from the rest energy

So the average energy density difference from the rest energy, relative to the rest energy, is

From eq. 1.0.24 and eq. 1.0.26 we have

or

This ratio of integrals is supposed to resolve to 1 and 2 in the low and high density limits. To consider this let’s perform one final non-dimensionalization, writing

The density, pressure, and energy take the form

We can rewrite the square roots in the number density and energy density expressions by expanding out the completion of the square

Expanding the distribution about , we have

allowing us to write, in the low density limit with respect to

Low density result

An exact integration of the various integrals above is possible in terms of special functions. However, that attempt (included below) introduced an erroneous extra factor of . Given that this end result was obtained by tossing all but the lowest order terms in and , let’s try that right from the get go.

For the pressure we have an integrand containing a factor

Our pressure, to lowest order in and is then

Our energy density to lowest order in and from eq. 1.0.33c is

Comparing these, we have

or in this low density limit

High density limit

For the high density limit write , so that the distribution takes the form

This can be approximated by a step function, so that

With a change of variables , we have

Comparing both we have

or

Wow. That’s pretty low!

Pressure integral

Of these the pressure integral is yields directly to Mathematica

where is a modified Bessel function [5] of the second kind as plotted in fig. 1.2.

Fig 1.2: Modified Bessel function of the second kind

Plugging this into the series for the pressure, we have

Plotting the summands for in fig. 1.4 shows that this mix of exponential Bessel and quadratic terms decreases with .

Plotting this sum in fig. 1.3 numerically to 10 terms, shows that we have a function that appears roughly polynomial in and .

Fig 1.3: Pressure to ten terms in z and theta

Fig 1.4: Pressure summands

For small it can be seen graphically that there is very little contribution from anything but the term of this sum. An expansion in series for a few terms in and gives us

This allows a and approximation of the pressure

Number density integral

For the number density, it appears that we can evaluate the integral using integration from parts applied to eq. 1.0.30.30

Expanding in series, gives us

Here the binomial coefficient has the meaning given in the definitions of \statmechchapcite{nonIntegralBinomialSeries}, where for negative integral values of we have

Expanding in series to a couple of orders in and we have

To first order in and this is

which allows a relation to pressure

It’s kind of odd seeming that this is quadratic in temperature. Is there an error?

Energy integral

Starting from eq. 1.0.30c and integrating by parts we have

The integral with the factor of doesn’t have a nice closed form as before (if you consider the a nice closed form), but instead evaluates to a confluent hypergeometric function [6]. That integral is

and looks like fig. 1.5. Series expansion shows that this hypergeometricU function has a singularity at the origin

Fig 1.5: Plot of HypergeometricU, and with theta^5 scaling

so our multiplication by brings us to zero as seen in the plot. Evaluating the complete integral yields the unholy mess

to first order in and this is

Comparing pressure and energy we have for low densities (where )

or

It appears that I’ve picked up an extra factor of somewhere, but at least I’ve got the low density expression. Given that I’ve Taylor expanded everything anyways around and this could likely have been done right from the get go, instead of dragging along the messy geometric integrals. Reworking this part of this problem like that was done above.

Disclaimer

Question: Maximum entropy principle

Consider the “Gibbs entropy”

where is the equilibrium probability of occurrence of a microstate in the ensemble.

For a microcanonical ensemble with configurations (each having the same energy), assigning an equal probability to each microstate leads to . Show that this result follows from maximizing the Gibbs entropy with respect to the parameters subject to the constraint of

(for to be meaningful as probabilities). In order to do the minimization with this constraint, use the method of Lagrange multipliers – first, do an unconstrained minimization of the function

then fix by demanding that the constraint be satisfied.

For a canonical ensemble (no constraint on total energy, but all microstates having the same number of particles ), maximize the Gibbs entropy with respect to the parameters subject to the constraint of

(for to be meaningful as probabilities) and with a given fixed average energy

where is the energy of microstate . Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

then fix by demanding that the constraint be satisfied. What is the resulting ?

For a grand canonical ensemble (no constraint on total energy, or the number of particles), maximize the Gibbs entropy with respect to the parameters subject to the constraint of

(for to be meaningful as probabilities) and with a given fixed average energy

and a given fixed average particle number

Here represent the energy and number of particles in microstate . Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

then fix by demanding that the constrains be satisfied. What is the resulting ?

Answer

Writing

our unconstrained minimization requires

Solving for we have

The probabilities for each state are constant. To fix that constant we employ our constraint

or

Inserting eq. 1.15 fixes the probability, giving us the first of the expected results

Using this we our Gibbs entropy can be summed easily

or

For the “action” like quantity that we want to minimize, let’s write

for which we seek , such that

or

Our probability constraint is

or

Taking logs we have

We could continue to solve for explicitly but don’t care any more than this. Plugging back into the probability eq. 1.0.16 obtained from the unconstrained minimization we have

or

To determine we must look implicitly to the energy constraint, which is

or

The constraint () is given implicitly by this energy constraint.

Again write

The unconstrained minimization requires

or

The unit probability constraint requires

or

Our probability is then

The average energy and average number of particles are given by

The values and are fixed implicitly by requiring simultaneous solutions of these equations.

Question: Fugacity expansion ([3] Pathria, Appendix D, E)

The theory of the ideal Fermi or Bose gases often involves integrals of the form

where

denotes the gamma function.

Obtain the behavior of for keeping the two leading terms in the expansion.

For Fermions, obtain the behavior of for again keeping the two leading terms.

For Bosons, we must have (why?), obtain the leading term of for .

Answer

For we can rewrite the integrand in a form that allows for series expansion

For the th power of in this series our integral is

Putting everything back together we have for small

We’ll expand about , writing

The integral has been split into two since the behavior of the exponential in the denominator is quite different in the ranges. Observe that in the first integral we have

Since this term is of order 1, let’s consider the difference of this from , writing

or

This gives us

Now let’s make a change of variables in the first integral and in the second. This gives

As gets large in the first integral the integrand is approximately . The exponential dominates this integrand. Since we are considering large , we can approximate the upper bound of the integral by extending it to . Also expanding in series we have

For the remaining integral, Mathematica gives

where for

This gives

or

or

Evaluating the numerical portions explicitly, with

so to two terms (), we have

In order for the Boson occupation numbers to be non-singular we require less than all . If that lowest energy level is set to zero, this is equivalent to . Given this restriction, a substitution is convenient for investigation of the case. Following the text, we'll write

For , this is integrable

so that

Taylor expanding we have

Noting that , we have for the limit

or

For values of , the denominator is

To first order this gives us

Of this integral Mathematica says it can be evaluated for , and has the value

From [1] 6.1.17 we find

with which we can write

Question: Nuclear matter ([2], prob 9.2)

Consider a heavy nucleus of mass number . i.e., having total nucleons including neutrons and protons. Assume that the number of neutrons and protons is equal, and recall that each of them has spin- (so possessing two spin states). Treating these nucleons as a free ideal Fermi gas of uniform density contained in a radius , where , calculate the Fermi energy and the average energy per nucleon in MeV.

Answer

Our nucleon particle density is

With for the mass of either the proton or the neutron, and , the Fermi energy for these particles is

With , and for either the proton or the neutron, this is

This gives us

In lecture 16

we found that the total average energy for a Fermion gas of particles was

so the average energy per nucleon is approximately

Question: Neutron star ([2], prob 9.5)

Model a neutron star as an ideal Fermi gas of neutrons at moving in the gravitational field of a heavy point mass at the center. Show that the pressure obeys the equation

where is the gravitational constant, is the distance from the center, and is the density which only depends on distance from the center.

Answer

In the grand canonical scheme the pressure for a Fermion system is given by

The kinetic energy of the particle is adjusted by the gravitational potential

Differentiating eq. 1.75 with respect to the radius, we have

Noting that is the average density of the particles, presumed radial, we have

Question: Diatomic molecule ([1] pr 4.7)

Consider a classical system of non-interacting, diatomic molecules enclosed in a box of volume at temperature . The Hamiltonian of a single molecule is given by

Study the thermodynamics of this system, including the dependence of the quantity on .

Answer

Partition function
First consider the partition function for a single diatomic pair

Now we can make a change of variables to simplify the exponential. Let’s write

or

Our volume element is

It wasn’t obvious to me that this change of variables preserves the volume element, but a quick Jacobian calculation shows this to be the case

Our remaining integral can now be evaluated

Our partition function is now completely evaluated

As a function of and as in the text, we write

Gibbs sum

Our Gibbs sum, summing over the number of molecules (not atoms), is

or

The fact that we can sum this as an exponential series so nicely looks like it’s one of the main advantages to this grand partition function (Gibbs sum). We can avoid any of the large approximations that we have to use when the number of particles is explicitly fixed.

Pressure

The pressure follows

Average energy

or

Average occupancy

but this is just , or

Free energy

Entropy

Expectation of atomic separation

The momentum portions of the average will just cancel out, leaving just

Question: Quantum anharmonic oscillator ([1] pr 3.30)

The energy levels of a quantum-mechanical, one-dimensional, anharmonic oscillator may be approximated as

The parameter , usually , represents the degree of anharmonicity. Show that, to the first order in and the fourth order in , the specific heat of a system of such oscillators is given by

Answer

We can expand the partition function in a first order Taylor series about , then evaluate the sums

The quadratic sum can be evaluated indirectly as it can be expressed as a derivative

Finally, evaluation of the derivatives gives us

Now we’d like to compute the specific heat in terms of derivatives of . First, for the average energy

The specific heat is

or

Actually computing that is messy algebra (See \nbref{pathria_3_30.nb}), and the result isn’t particularily interesting looking. The plot fig. 1.1 is interesting though and shows negative heat capacities near zero and a funny little jog near .

Fig 1.1: Quantum anharmonic heat capacity

Also confirmed in the Mathematica notebook is equation eq. 1.0.2, which follows by first doing a first order series expansion in , then a subsequent series expansion in .

Consider a single particle perturbation of a classical simple harmonic oscillator Hamiltonian

Calculate the canonical partition function, mean energy and specific heat of this system.

This problem can be attempted in two ways, the first of which was how I did it on the midterm, differentiating under the integral sign, leaving the integrals in exact form, but not evaluated explicitly in any way.

Alternately, by Taylor expanding around and with those as the variables in the Taylor expansion (as now done in the Pathria 3.29 problem), we can form a solution in short order. Given my low midterm mark, it seems very likely that this was what was expected.

This problem was suggested as prep for the second midterm, but I spent too much time on my problem set. That’s pretty unfortunate since this showed exactly the approach that was expected for the second midterm problem. Not hard, just not obvious in the heat of the moment how to do that Taylor expansion.

Question: Anharmonic oscillator ([1] pr 3.29)

The potential energy of a one-dimensional, anharmonic oscillator may be written as

where , , and are positive constant; quite generally, and may be assumed to be very small in value.

Show that the leading contribution of anharmonic terms to the heat cpacity of the oscillator, assumed classical, is given by

To the same order, show that the mean value of the position coordinate is given by

Answer

Our partition function is

How to expand this wasn’t immediately clear to me (as it wasn’t on the midterm either). We can’t Taylor expand in , because there’s no single position that is of interest to expand around (we are integrating over all ). What we can do though is Taylor expand about the values and , which are assumed to be small. Here’s the two variable Taylor expansion of this perturbed harmonic oscillator exponential. With

The expansion to second order is

This can now be integrated by parts, where any odd powers are killed. For even powers we have

This gives us

Retaining only the first two terms of the expansion, we have

Our average energy, in this approximation

So to first order in our specific heat is

or

For the coordinate

or

Compare this to the expectation of the coordinate for an unperturbed harmonic oscillator

We now have a temperature dependence to the expecation of the coordinate that we didn’t have for the harmonic oscillator.

Here’s some reflection about this Thursday’s midterm, redoing the problems without the mad scramble. I don’t think my results are too different from what I did in the midterm, even doing them casually now, but I’ll have to see after grading if these solutions are good.

Question: Magnetic field spin level splitting (2013 midterm II p1)

A particle with spin has states . When exposed to a magnetic field, state splitting results in energy . Calculate the partition function, and use this to find the temperature specific magnetization. A “sum the geometric series” hint was given.

Answer

Our partition function is

Writing

that is

Substitution of gives us

To calculate the magnetization , I used

As [1] defines magnetization for a spin system. It was pointed out to me after the test that magnetization was defined differently in class as

These are, up to a sign, identical, at least in this case, since we have and travelling together in the partition function. In terms of the average energy

Compare this to the in-class definition of magnetization

For this derivative we have

This gives us

After some simplification (done offline in \nbref{midtermTwoQ1FinalSimplificationMu.nb}) we get

I got something like this on the midterm, but recall doing it somehow much differently.